JEE Main 202226 Jul, Morning shiftPhysicsEasyElectromagnetic Waves120× since 2002The magnetic field of a plane electromagnetic wave is given by : B→=2×10−8sin(0.5×103x+1.5×1011t) j^ T\overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \,\hat{j} \mathrm{~T}B=2×10−8sin(0.5×103x+1.5×1011t)j^ T. The amplitude of the electric field would be :AOption A. 6 Vm−16\, \mathrm{Vm}^{-1}6Vm−1 along xxx-axisBOption B. 3 Vm−13\, \mathrm{Vm}^{-1}3Vm−1 along zzz-axisCOption C. 6 Vm−16\, \mathrm{Vm}^{-1}6Vm−1 along zzz-axisDOption D. 2×10−8 Vm−12 \times 10^{-8} \,\mathrm{Vm}^{-1}2×10−8Vm−1 along zzz-axisCheckSkip · show solution
